The reaction between phosphoric acid and ammonia that produces ammonium phosphate can be written as follows:
3NH3 + H3PO4 ..................> (NH4)3PO4
From the periodic table:
molar mass of nitrogen = 14 grams
molar mass of hydrogen = 1 grams
molar mass of oxygen = 16 grams
molar mass of phosphorus = 30.9 grams
based on this:
molar mass of 3NH3 = 3 (14 + 3(1)) = 51 grams
molar mass of H3PO4 = 3(1) + 30.9 + 4(16) = 97.9 grams
molar mass of (NH4)3PO4 = 3 (14 + 4(1)) + 30.9 + 4(16) = 54 + 30.9 + 64
= 148.9 grams
Therefore, 97.9 grams of phosphoric acid is required to produced 148.9 grams of ammonium phosphate.
Thus, to know the mass of ammonium phosphate produced from 4.9 grams of phosphoric acid, we will simply use cross multiplication as follows:
amount of produced ammonium phosphate = (4.9 x 148.9) / 97.9 = 7.45 g
Answer:
See explanation
Explanation:
According to Louis de Broglie, matter has an associated wavelength. He was the first scientist to establish the idea of wave-particle duality or wave- particle paradox.
The display of wavelike properties by objects in the universe is dependent on the magnitude of the of the mass of the body. Small objects have a large associated wavelength and can be described completely by quantum mechanics.
A buckyball with a mass of 1.2 x 10-21 g, 0.7 nm wide, moving at 38. m/s has a very small mass and significant associated wavelength hence the system can be completely described by quantum mechanics.
Balance the equation: 2Na + S --> Na2S
Using the given amount of the reactants in the reaction, calculate the amount of the product:
45.3g Na x (1 mol/22.99 g)= 1.97 mol of Na
105f S x (1 mol/ 32.06g) = 3.28 mol of S
The limiting reactant would be Na:
<span>1.97 mol Na x (1 mol Na2S/ 2 mol Na) x (78.04g/mol) = 76.87g of Na2S produced</span>
Answer:
Explanation:
From the given information:
Camphor may be reduced as readily in the presence of sodium borohydride(NaHB4). The resulting compound which is stereoselective requires 1 mole of sodium borohydride (NaHB4) to reduce 1 mole of camphor in this reaction. The reaction is shown below.
Through the reduction process of camphor, the reducing agent can reach the carbonyl face with a one-carbon linkage. The product stereoisomer is known as borneol.
If the molecular weight of camphor = 152.24 g/mol
and it mass = 200 mg
The its no of moles = 200 mg/ 152.24 g/mol
= 1.3137 mmol
Now the amount of the required mmol for NaBH4 to be consumed in the reaction = 5.2 × 1.3137 mmol
= 6.831 mmol
since the molar mass of NaBH4 = 37.83 g/mol
Then, using the same formula:
No of moles = mass/molar mass
mass = No of moles × molar mass
mass = 6.831 mmol × 37.83 g/mol
mass of NaBH4 used = 258.42 mg
